Re: [Boost-users] transform_view: question about boilerplate code

Zeljko Vrba wrote:

I have succeeded in writing the following transform functor:

typedef bf::map, bf::pair > Map; typedef bf::vector Pair;

struct Map2Pair { template<typename Sig> struct result;

template<typename U> struct result : remove_reference { };

template<typename T> typename T::second_type operator()(T t) const { return t.second; } };

I have tested it, it works. The result magic has been copied from the transform_view example in the documentation and adjusted for the proper return type. Where can I read a bit more about result conventions?

Read on Boost.Result of and the TR1 result_of conventions. http://www.boost.org/doc/libs/1_36_0/libs/utility/utility.htm http://anubis.dkuug.dk/jtc1/sc22/wg21/docs/papers/2003/n1454.html In particular,

how does the

template<typename U> struct result : remove_reference { };

specialization work when Map2Pair is a struct, and it doesn't even have a constructor!

It's not a function call. It's a function declarator. This is first discussed and put into use in the Boost.Function library: http://www.boost.org/doc/libs/1_36_0/doc/html/function/tutorial.html#id29033...

Even worse, why does it work at all when T is not in the scope of the template (.. because it's never used?)

Which template? Yes, it is not used,

the following functor also works:

struct Map2Pair { template<typename T> typename T::second_type operator()(T t) const { return t.second; } };

When and how is the above version preferred than the simpler version?

Are you sure the one above works? You need the result type, AFAICT. Regards, -- Joel de Guzman http://www.boostpro.com http://spirit.sf.net